Physics 68 Lagrangian Mechanics (19 of 32) Oscillating Bar
Summary
TLDREn esta conferencia en línea, se analiza el movimiento de una barra articulada en un extremo y sostenida por una resortera en el otro. La barra oscila alrededor de un equilibrio horizontal debido a la interacción entre la fuerza de gravedad y la tensión del resorte. El objetivo es encontrar la ecuación del movimiento utilizando el método de Lagrange, calculando la energía cinética y potencial del sistema. Se resuelve la ecuación de la barra en equilibrio y se sustituye en las expresiones para obtener la ecuación de movimiento final, que describe las oscilaciones de la barra.
Takeaways
- 😀 La clase trata sobre el movimiento de una barra articulada en un extremo y soportada por una muelle en el otro.
- 🔧 La barra puede oscilar alrededor de su posición horizontal de equilibrio tras ser desplazada y soltada.
- 🌀 Se utiliza el método de Lagrange para encontrar la ecuación del movimiento de este sistema mecánico.
- 📚 Se definen las energías cinética y potencial del sistema, que son fundamentales para la aplicación del método de Lagrange.
- 🏋️ La energía cinética del sistema es dada por la mitad del momento de inercia multiplicado por la velocidad angular al cuadrado.
- ⚖️ El momento de inercia de la barra es un tercio de la masa multiplicada por la longitud al cuadrado.
- 📉 La energía potencial incluye la energía almacenada en la muelle y la energía potencial gravitacional perdida por la barra.
- 🔍 Se determina la constante de elongación inicial de la muelle (x_nought) a partir de la condición de equilibrio de torques.
- 📐 La ecuación de movimiento se obtiene tomando derivadas parciales del Lagrangiano con respecto a theta y theta punto (velocidad angular y aceleración angular).
- 🧩 Al final, se simplifica la ecuación para obtener la forma más compacta que describe el movimiento oscilatorio de la barra: θ'' + (3k/m)θ = 0.
- 🤔 El guion sugiere que el proceso de derivación puede ser complejo y requiere atención para evitar errores.
Q & A
¿Qué tipo de sistema se está analizando en la conferencia?
-Se está analizando un sistema de una barra que puede girar en un extremo y está soportada por una muelle en el otro extremo.
¿Cuál es la condición inicial de la muelle en el sistema descrito?
-Inicialmente, la muelle no alcanza todo el camino hasta la barra, lo que significa que está en su longitud natural.
¿Cómo se define la variable 'x' en el contexto de la conferencia?
-La variable 'x' se refiere a la extensión adicional de la muelle cuando se estira más allá de su longitud natural para alcanzar la barra horizontal.
¿Qué variable se utiliza para describir el movimiento de la barra en el sistema?
-La variable utilizada para describir el movimiento de la barra es el ángulo theta (θ).
¿Cómo se calcula la energía cinética del sistema?
-La energía cinética se calcula como la mitad del momento de inercia multiplicado por la velocidad angular al cuadrado (θ˙²), donde el momento de inercia para una barra en un extremo es un tercio de la masa multiplicada por el cuadrado de la longitud de la barra.
¿Cuál es la fórmula para el momento de inercia de una barra en un extremo?
-El momento de inercia de una barra en un extremo es un tercio de la masa (m) multiplicada por el cuadrado de la longitud (l) de la barra.
¿Cómo se determina la energía potencial del sistema?
-La energía potencial se determina como la energía potencial almacenada en la muelle más la energía potencial perdida debido a la caída del centro de masa de la barra.
¿Cómo se relaciona la extensión de la muelle con la posición de la barra?
-La extensión de la muelle se relaciona con la posición de la barra a través de la suma de la extensión natural 'x' más la extensión adicional 'lθ', donde 'l' es la longitud de la barra y 'θ' es el ángulo de rotación.
¿Cómo se establece el equilibrio entre la fuerza de gravedad y la fuerza de la muelle en la barra horizontal?
-El equilibrio se establece cuando la fuerza de gravedad en el centro de masa de la barra es igual y opuesta a la fuerza de la muelle en su estado natural, lo que se expresa como 'kx = mg/2'.
¿Cuál es la ecuación final del movimiento para el sistema de la barra y la muelle?
-La ecuación del movimiento final es θ¨ + (3k/m)θ = 0, donde θ¨ representa la aceleración angular y 'k' y 'm' son la constante de la muelle y la masa de la barra, respectivamente.
Outlines
😀 Introducción a la oscilación de una barra con resorte
El primer párrafo presenta un sistema físico compuesto por una barra que puede girar sobre un extremo y está soportada por un resorte en el otro. Se describe cómo la barra, inicialmente horizontal, se oscila debido al equilibrio entre la fuerza de gravedad y la tensión del resorte. El objetivo es encontrar la ecuación del movimiento de este sistema utilizando la variable ángular theta. Se introduce la constante 'x de no', que representa la extensión adicional del resorte más allá de su longitud natural, y se discuten las fórmulas para la energía cinética y potencial del sistema, utilizando el momento de inercia y la ecuación de la energía potencial del resorte.
🔍 Análisis de la energía cinética y potencial del sistema
En el segundo párrafo, se profundiza en el cálculo de la energía cinética y potencial del sistema. Se establece que la energía cinética es proporcional al momento de inercia, que para una barra en rotación es uno tercero de la masa multiplicada por el cuadrado de la longitud. La energía potencial se compone de la energía almacenada en el resorte, que es proporcional al cuadrado de su extensión, y la pérdida de energía potencial debido a la caída del centro de masa de la barra. Se resuelve la constante 'x de no' utilizando el equilibrio de torques y se sustituye en la fórmula de energía potencial para obtener expresiones en términos de variables conocidas.
📚 Aplicación del método de Lagrange para hallar la ecuación del movimiento
El tercer párrafo detalla el uso del método de Lagrange para derivar la ecuación del movimiento de la barra. Se calcula el Lagrangiano como la diferencia entre la energía cinética y potencial, y se derivan parcialmente con respecto a theta y theta punto para encontrar las ecuaciones de movimiento. Se simplifica la expresión para obtener una ecuación de segunda orden en términos de theta doble punto, que representa la aceleración angular de la barra. Finalmente, se presenta la ecuación del movimiento en su forma más compacta, mostrando que es una oscilación armónica con una frecuencia determinada por las constantes del sistema.
Mindmap
Keywords
💡Lectura
💡Palabra clave
💡Barra articulada
💡Resorte
💡Equilibrio
💡Oscilación
💡Ecuación del movimiento
💡Lagrangiano
💡Energía cinética
💡Energía potencial
💡Momento de inercia
💡Derivada parcial
💡Torque
💡Elongación
Highlights
Introduction to a lecture on the dynamics of a hinged bar supported by a spring.
Description of the bar's initial state and the natural length of the spring.
Explanation of the equilibrium state when the bar is horizontal.
Introduction of the variable 'x' representing the extra extension of the spring.
Use of the bar's length 'l' and angle 'theta' to calculate the spring's elongation.
Introduction of the Lagrangian method to solve the system's equation of motion.
Calculation of the kinetic energy of the system considering rotational motion.
Derivation of the moment of inertia for the bar and its relation to mass and length.
Expression of the kinetic energy in terms of variable 'theta'.
Complex calculation of the system's potential energy involving spring force and gravitational potential.
Identification of the constant 'x_not' and its relation to the spring's natural length.
Derivation of the torque balance equation at the equilibrium state.
Calculation of 'x_not' using the torque balance and given constants.
Substitution of 'x_not' into the potential energy equation.
Formulation of the Lagrangian with kinetic and potential energy components.
Differentiation of the Lagrangian to find the equation of motion.
Final equation of motion for the oscillating bar and spring system.
Verification of the derived equation and its significance in understanding the system's dynamics.
Transcripts
welcome to our lecture online
the examples are beginning to be more
and more interesting here so what we
have here is a bar
which can hinge on one end
and it's supported by spring on the
other end
now initially the spring would not reach
all the way down to the bar this would
be the natural length of the spring so
when you attach
the spring to the bar
the weight of the bar will pull the bar
will pull the spring until the force of
gravity on the bar causing it well
actually the torque on the bar and the
torque caused by the spring until those
are at equilibrium and supposedly there
will be equilibrium when the bar is
horizontal so normally if you place the
bar in the horizontal state attached to
the spring then nothing would move we
would just sit there and then of course
once you pull it down and you let go the
bar would oscillate back and forth and
we're trying to find the equation of
motion of this particular system
let's call the distance from where the
unstretched spring reaches to where the
bar is horizontal that extra extension
of the spring where we pull it down then
let's call that x of not that's not a
variable that is actually a constant
value
and we don't know yet what that is
and then the additional distance of
spring is elongated notice we can use
l the length of the bar times the angle
theta to get that additional distance
right there so it looks like
the variable that we're working with
will be the variable theta
now what we need to do to solve this
using the lagrangian we'll need to find
the kinetic energy and the potential
energy of the system and so let's start
with the kinetic energy of the system
the kinetic energy
will be equal to and now notice that the
bar will only swing back and forth like
this attached to one end so we're
looking at rotational motion of the bar
so the kinetic energy will be one half
the moment of inertia times omega
squared
well the moment of inertia of a bar
that's attached to one end is one-third
the length times
well we also need the mass let me put
the mass it would be one-third the mass
times the length squared so this becomes
equal to one-half times one-third the
mass times the length squared and we'll
we'll write the l as a little l to that
to differentiate it from the big l for
the lagrangian so that's the moment of
inertia and then omega well omega that
would be theta dot so we'll write as
theta dot squared
and that means that the kinetic energy
then becomes 1 6
m
l squared
theta dot squared and so that will then
be the kinetic energy in terms of the
only variable in the system theta and
that's why we're only looking for a
single equation of motion
now we need the potential energy and
that's a little bit more tricky here
okay let's try it
so the potential energy
is equal to well first of all it'll be
the potential energy of the
spring
and that means it's going to be one half
k
times the elongation squared and the
elongation will be the sum of x sub
naught plus
the l theta quantity squared
that will be the potential energy stored
in the spring and then minus because the
bar loses height now the center mass
will be at the middle so the distance
that the middle drops would be
half of this so the drop in potential
energy will be m times g times half of l
theta so l theta over 2
and that will then give us the potential
energy
but notice we're stuck with this x sub
naught we know it's a constant but what
is that constant equal to
well notice that we said in the
beginning that if everything is just in
a constant stationary mode in a static
mode with the bar horizontal the weight
of the bar will be canceled out by the
force on the spring keeping the bar
horizontal which means that the sum of
the torques about this point must add up
to zero so the sum
of the torques add up to zero and so
what does that equal to well we have the
mg and let me use a different color
so we have this torque right here that
mg and notice that gives us a clockwise
torque which is a negative torque that
would be minus
m
g
times the distance and that distance
here well since the angles are very
small we can simply call that that
distance half l so it would be l over 2
l over 2
and that would be the force times the
perpendicular distance from the point of
rotation to the line of action the force
it's negative because it gives you a
clockwise torque and then minus
the force of the spring
and let's see here the force of the
spring that would be in its natural
unstretched state well stretch because
the spring and natural length is this
but without pulling it down any further
from its equilibrium point and so that
would be kx
so kx
sub naught times l
because we have to multiply the force
times the perpendicular distance and so
we get l
all right so now this balances out so we
have the torque caused by the weight of
the beam
through the center mass counterbalance
oh uh let's see here that would be plus
i'm sorry that would be a plus because
this acts in a counterclockwise
clockwise direction so it's plus
the force on the spring kx times the
distance from where this force acts to
the point of rotation
and since that's equal to zero that
means that the l's cancel out
like this
which means that we can write this as kx
sub naught
is equal to when we move the other side
mg over 2 and i'm moving the k down here
so x sub naught equals
m g over 2 k
which means we we can replace except not
by mg over 2k and substitute that in
here so the potential energy
is equal to one-half k
times
mg
over
2k
plus
l theta quantity squared minus
mg
l over 2. oh l theta over 2.
all right so now we have the kinetic
energy
and we have the potential energy of the
system
and we've replaced x sub not which is a
constant by
variables or constants that we know we
know m g and k supposedly they gave
those to us
now we're ready to find this equation so
first we find the potential the partial
derivative
of l well
before i do that i should write what l
is equal to right so l is equal to the
kinetic energy minus the potential
energy so this is equal to the kinetic
energy of 1 6
m
l squared
theta dot squared
minus this so that becomes minus one
half k
times
mg
over 2k
plus
l theta quantity squared
and then minus times a minus gives us a
plus
mg
l theta over 2. all right so now we have
the lagrangian and now we can find the
partial of the lagrangian with respect
to
theta dot
x equal to notice there's no theta dot
over this part only over here so we have
2 times 1 6 which is 1 3
ml squared
times
theta
dot
okay and now we're going to take the
time derivative of this so now we take
the d dt
of the partial of the lagrange with
respect to theta dot
and so that simply gives us 1 3
m l squared theta double dot or the
acceleration of that
so
there we have this part of the
equation now we need to find the partial
value with respect to x oh
well in this case of course it's going
to be theta right so we find the partial
value with respect to theta
and that means that this becomes zero
this part becomes zero but here's the
theta in there and there's a theta as
well all right
so
the partial of l with respect to theta
is equal to
so we have minus one half k times oh
wait a minute this thing here is squared
hmm
well you know what
to make it easier we may want to just
multiply all that out
well let's try that let's try that so
i'm going to repeat l over here but
multiply this part out here that just
makes it easier to see otherwise it may
be too too easy to make a mistake so
let's try that so we write l is equal to
1 6
m
l squared theta dot squared
minus one half k
times
the first quantity squared so we get m
g over 2k quantity squared that's just a
constant squared
plus twice the product of what's inside
so plus
2 times the product that will cancel out
this 2 right here so we end up mgk l
theta mg over kl theta mg over k
times l times theta
so twice the product of those two that
gets rid of the two and this squared so
plus
l squared theta squared
and notice that's multiplied by times
the minus one half k
into those terms and then at the end we
still have a plus
m g
l theta over two and now when we do this
it makes it a lot easier to take care of
that so that goes to zero the first term
is a constant that goes to zero here we
have a theta in there so it is minus one
half k times this the k's cancel out we
end up with uh
well we still have the minus don't we we
still have the minus so minus
the k's cancel
mgl
over 2.
and then here
we end up with
2 times a half that twos the twos cancel
we have a minus so we have minus
we have k
l squared
theta to the first power
the 2 is gone all right
and then we take the partial of this
or the partial of this with respect to
theta so we end up with a minus or plus
i should say mgl
over 2.
well notice this i have a minus mgl over
2 and a plus mgl over 2 so that cancels
out
that's always nice when we can do that
and we're left with a minus kl squared
theta
for the partial of the lagrangian with
respect to theta and now finally
when we want to find the equation of
motion we take the first part which is
this so we end up with 1 3
m
l squared theta double dot
minus
the partial value with respect to in
this case theta which is minus this that
becomes plus
k l squared theta and that equals zero
and of course you could
rewrite it where you have theta double
dot in the front
so that means oh well wait a minute here
we can't simplify the l squared cancels
out so that makes it nice l squared
cancels out
we multiply everything by 3 and divide
everything by m so essentially what we
can write this as theta double dot
plus 3k
over m
theta equals zero and maybe that's a
more compact form to write the final
answer and that's then the equation of
motion of that oscillating bar attach
that spring and that
is how it's done
i think you fell asleep
let's see if i got this right
yeah
you
Ver Más Videos Relacionados
Movimiento armónico simple | El péndulo simple
Physics 68 Lagrangian Mechanics (18 of 32) Two Mass - Two Spring System
Understanding How Torque Works
Conservación de la energía y resorte vertical | Física | Khan Academy en Español
Modelado matemático de péndulo simple
Cinemática 3D: Ecuación del Movimiento Rectilíneo Uniforme (MRU)
5.0 / 5 (0 votes)